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3_sincbasis.tex
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\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\begin{document}
\begin{frame} \frametitle{Overview:}
\begin{itemize}
\item Spectrum of interpolated signals
\item Space of bandlimited functions
\item Sinc sampling
\item The sampling theorem
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Sinc interpolation}
the ingredients:
\begin{itemize}
\item discrete-time signal $x[n]$, $n\in\mathbb{Z}$ (with DTFT $X(e^{j\omega})$)
\item interpolation interval $T_s$
\item the sinc function
\end{itemize}
\vspace{2em}
\pause
the result:
\begin{itemize}
\item a smooth, continuous-time signal $x(t)$, $t\in\mathbb{R}$
\end{itemize}
\vspace{2em}
\pause
\centering
what does the spectrum of $x(t)$ look like?
\end{frame}
\begin{frame} \frametitle{Key facts about the sinc}
\centering
\begin{tabular}{ccc}
$\displaystyle \varphi(t) = \sinc\left(\frac{t}{T_s}\right)$
&
$\longleftrightarrow$
&
$\displaystyle\Phi(f) = \frac{1}{F_s} \rect\left(\frac{f}{F_s}\right)$
\\ \\ \\
&
$\displaystyle T_s = \frac{1}{F_s}$
&
\end{tabular}
\end{frame}
\begin{frame} \frametitle{Key facts about the sinc}
\centering
\begin{figure}
\begin{dspPlot}[xticks=custom,height=2cm,xout=true,sidegap=0,ylabel={$\varphi(t)$}]{-10,10}{-0.3,1.1}
\moocStyle
\dspFunc{x \dspSinc{0}{1}}
\dspCustomTicks[axis=x]{0 0 1 $T_s$}
\end{dspPlot}
\end{figure}
\begin{figure}
\begin{dspPlot}[xticks=custom,yticks=custom,height=2cm,sidegap=0,ylabel={$\Phi(f)$}]{-3,3}{0,1.2}
\moocStyle
\dspFunc{x \dspRect{0}{1}}
\dspCustomTicks[axis=x]{0 0 0.5 $F_s/2$}
\dspCustomTicks[axis=y]{0 0 1 $1/F_s$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Sinc interpolation}
\[
x(t) = \sum_{n = -\infty}^{\infty}x[n]\,\mbox{sinc}\left(\frac{t - nT_s}{T_s}\right)
\]
\end{frame}
\begin{frame} \frametitle{Spectral representation (I)}
\begin{align*}
X(f) &= \int_{-\infty}^{\infty} x(t)\, e^{-j2\pi f t}dt \\ \pause
&= \int_{-\infty}^{\infty} \sum_{n = -\infty}^{\infty}x[n]\sinc\left(\frac{t - nT_s}{T_s}\right) e^{-j2\pi f t}dt \\ \pause
&= \sum_{n = -\infty}^{\infty} x[n] \int_{-\infty}^{\infty} \sinc\left(\frac{t - nT_s}{T_s}\right) e^{-j2\pi f t}dt \\ \pause
&= \sum_{n = -\infty}^{\infty} x[n] \left(\frac{1}{F_s}\right) \rect\left(\frac{f}{F_s}\right) \, e^{-j2\pi f\, n T_s}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Spectral representation (II)}
\begin{align*}
X(f) &= \sum_{n = -\infty}^{\infty} x[n] \left(\frac{1}{F_s}\right) \rect\left(\frac{f}{F_s}\right) \, e^{-j2\pi f \, n T_s} \\ \pause
&= T_s \rect\left(\frac{f}{F_s}\right) \sum_{n = -\infty}^{\infty} x[n] e^{-j2\pi(f/F_s) n} \\ \\ \pause
&= \begin{cases}
T_s\, X(e^{j2\pi f/F_s}) & \mbox{for $|f| \leq F_s/2$} \\
0 & \mbox{otherwise}
\end{cases}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Spectral representation (III)}
\[
X(f) = \begin{cases}
T_s\, X(e^{j2\pi f/F_s}) & \mbox{for $|f| \leq F_s/2$} \\
0 & \mbox{otherwise}
\end{cases}
\]
\vspace{2em}
\begin{itemize}
\item map $\omega = \pi$ to $f = F_s/2$
\item scale spectrum by $T_s$ (total energy constant)
\item rect keeps only the baseband copy of the periodic digital spectrum
\end{itemize}
\end{frame}
\def\plotSpec#1#2{\dspFunc[#2]{x \dspPorkpie{0}{#1} #1 div}}
\def\plotCurrent#1#2#3#4#5{%
\only<#1>{
\FPupn\o{#2 1 / 2 trunc}
\plotSpec{\o}{}
\dspCustomTicks[axis=x]{0 0 {-\o} #5 {\o} #4}
\dspText(-2,1.8){$T_s=$#3}}}
\def\plotPast#1#2{
\only<#1->{
\FPupn\o{#2 1 / 3 trunc}
\plotSpec{\o}{linecolor=gray}}}
\begin{frame} \frametitle{Spectrum of interpolated signals}
\centering
\begin{figure}
\begin{dspPlot}[xtype=freq,height=2cm,ylabel={$X(e^{j\omega})$}]{-1,1}{0,1.2}
\moocStyle
\dspFunc{x \dspPorkpie{0}{1}}
\end{dspPlot}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom,height=2cm,ylabel={$X(f)$}]{-2.5,2.5}{0,2.4}
\moocStyle
\plotCurrent{1}{1}{$T_0$}{$F_s/2=1/(2T_0)$}{$-F_s/2$}
\plotPast{2}{1}
\plotCurrent{2}{2}{$2T_0$}{$F_s/2=1/(4T_0)$}{$-F_s/2$}
\plotPast{3}{2}
\plotCurrent{3}{0.5}{$T_0/2$}{$F_s/2=1/T_0$}{$-F_s/2$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Spectrum of interpolated signals}
pick interpolation period $T_s$:
\begin{itemize}[<+->]
\item $X(f)$ is $F_s$-bandlimited, with $F_s = 1/T_s$
\item fast interpolation ($T_s$ small) $\rightarrow$ wider spectrum
\item slow interpolation ($T_s$ large) $\rightarrow$ narrower spectrum
\item (for those who remember...) it's like changing the speed of a record player
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Space of bandlimited functions}
\note<1>{here we ask the question of whether \\ we can invert the process. \\ Given an $\ell_2$ sequence we get a \\ unique BL function, can we get \\ a unique sequence from a BL function so \\ that we can get the BL function back \\ with sinc interpolation?}
\centering
\begin{tabular}{ccc}
&
$T_s$ &
\\
$x[n] \in \ell_2(\mathbb{Z})$ & \only<1>{$\xrightarrow{\hspace*{3cm}}$}\only<2>{{\color{darkred}$\xleftrightarrow{\hspace*{3cm}}$}} &
$x(t) \in L_2(\mathbb{R})$
\\
&
\only<2>{{\color{darkred}?}} &
$F_s$-BL
\end{tabular}
\end{frame}
\begin{frame} \frametitle{Let's lighten the notation}
\centering
for a while we will proceed with $T_s = 1$ (so that $F_s = 1$ as well)
(derivations in the general case are in the book)
\end{frame}
\begin{frame} \frametitle{The road to the sampling theorem}
claims:
\begin{itemize}[<+->]
\item the space of $1$-bandlimited functions is a Hilbert space
\item the functions $\varphi^{(n)}(t) = \sinc(t-n)$, with $n \in \mathbb{Z}$, form a basis for the space
\item if $x(t)$ is $1$-BL, the sequence $x[n] = x(n)$, with $n \in \mathbb{Z}$, is a sufficient representation\\ (i.e. we can reconstruct $x(t)$ from $x[n]$)
\end{itemize}
\end{frame}
\begin{frame} \frametitle{The space $1$-BL}
\begin{itemize}[<+->]
\item clearly a vector space because $1$-BL $\subset L_2(\mathbb{R})$ (and linear combinations of $1$-BL functions are $1$-BL functions)
\item inner product is standard inner product in $L_2(\mathbb{R})$
\item completeness... that's more delicate
\end{itemize}
\end{frame}
\begin{frame} \frametitle{The space of $1$-BL functions}
recap:
\begin{itemize}
\item inner product:
\[
\langle x(t), y(t) \rangle = \int_{-\infty}^{\infty} x^*(t)y(t) dt
\]
\item convolution:
\[
(x \ast y)(t) = \langle x^*(\tau), y(t-\tau) \rangle
\]
\end{itemize}
\end{frame}
\begin{frame} \frametitle{A basis for the $1$-BL space}
\note<1>{exploit the symmetry of the sinc \\ $\sinc(m-t) = \sinc(t-m)$ and \\ use the change of variable $\tau=t-n$. \\ ~\\ symmetry of the sinc will be used many times \\ later, so stress that}
\[
\varphi^{(n)}(t) = \sinc(t-n), \quad\quad n \in \mathbb{Z}
\]
\begin{align*}
\langle \varphi^{(n)}(t), \varphi^{(m)}(t) \rangle
&= \langle \varphi^{(0)}(t - n), \varphi^{(0)}(t - m) \rangle \\ \pause
&= \langle \varphi^{(0)}(t - n), \varphi^{(0)}(m-t) \rangle \\ \pause
&= \int_{-\infty}^{\infty} \sinc(t - n)\sinc(m-t)\, dt \\ \pause %\tau = t-n, t=\tau+n
&= \int_{-\infty}^{\infty} \sinc(\tau)\sinc((m-n)-\tau)\, d\tau \\ \pause
&= (\sinc \ast \sinc)(m-n)
\end{align*}
\end{frame}
\begin{frame} \frametitle{A basis for the $1$-BL space}
now use the convolution theorem knowing that:
\[
\FT{\sinc(t)} = \rect\left(f\right)
\]
\begin{align*}
(\sinc \ast \sinc)(m-n)
&= \int_{-\infty}^{\infty} \rect^2(f) \, e^{j2\pi f (m-n)} df \\ \pause
&= \int_{-1/2}^{1/2} e^{j2\pi f (m-n)}df \\ \pause
&= \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{j\Omega (m-n)}d\Omega \\ \pause
&= \begin{cases}
1 & \mbox{for $m=n$} \\
0 & \mbox{otherwise}
\end{cases}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Space of bandlimited functions}
\centering
\begin{tabular}{ccc}
$x[n] \in \ell_2(\mathbb{Z})$ & $\xleftrightarrow{\hspace*{3cm}}$ & $x(t) \in L_2(\mathbb{R})$
\\
& & $1$-BL
\end{tabular}
\end{frame}
\begin{frame} \frametitle{Sampling as a basis expansion}
\centering
for any $x(t) \in \mbox{$1$-BL}$:
\begin{align*}
\langle \varphi^{(n)}(t), x(t) \rangle
&= \langle \sinc(t - n), x(t) \rangle = \langle \sinc(n-t), x(t) \rangle \\ \pause
&= (\sinc \ast \, x)(n) \\ \pause
&= \int_{-\infty}^{\infty}\rect\left(f\right)X(f)e^{j2\pi f n} df\\ \pause
&= \int_{-\infty}^{\infty} X(f)e^{j2\pi f n} df \\ \pause
&= x(n)
\end{align*}
\end{frame}
\begin{frame} \frametitle{Sampling as a basis expansion, $1$-BL}
\centering
Analysis formula:
\[
x[n] = \langle \sinc(t - n), x(t) \rangle
\]
\vspace{2em}
Synthesis formula:
\[
x(t) = \sum_{n = -\infty}^{\infty}x[n]\,\sinc(t - n)
\]
\end{frame}
\begin{frame}
\frametitle{Sampling as a basis expansion, $F_s$-BL}
\centering
Analysis formula:
\[
x[n] = \langle \sinc\left(\frac{t - nT_s}{T_s}\right), x(t) \rangle = T_s\,x(nT_s)
\]
\vspace{2em}
Synthesis formula:
\[
x(t) = \frac{1}{T_s}\sum_{n = -\infty}^{\infty}x[n]\,\sinc\left(\frac{t - nT_s}{T_s}\right)
\]
\end{frame}
\begin{frame} \frametitle{The sampling theorem}
\begin{itemize}
\item the space of $F_s$-bandlimited functions is a Hilbert space
\item set $T_s = 1/F_s$
\item the functions $\displaystyle\varphi^{(n)}(t) = \sinc\left(\frac{t-nT_s}{T_s}\right)$ form a basis for the space %, with $n \in \mathbb{Z}$,
\item for any $x(t) \in \mbox{$F_s$-BL}$ the coefficients in the sinc basis are the (scaled) samples $T_s\,x(nT_s)$
\end{itemize}
\pause
\vspace{2em}
\centering
for any $x(t) \in \mbox{$F_s$-BL}$, a sufficient representation is the sequence $x[n] = x(nT_s)$
%with $T_s = \pi/\Omega_N$. %, $\quad n\in \mathbb{Z}$,
\end{frame}
\begin{frame} \frametitle{The sampling theorem, corollary}
\begin{itemize}
\item $F_s$-BL $\subseteq$ $F$-BL for any $F \ge F_s$
\end{itemize}
\pause
\vspace{2em}
\centering
for any $x(t) \in \mbox{$F_s$-BL}$, a sufficient representation is the sequence \\ $x[n] = x(nT_s)$ for any $T_s \le 1/F_s$
\end{frame}
\begin{frame}
\frametitle{The sampling theorem, again}
\centering
any signal $x(t)$ whose highest frequency component is $F_N$~Hz \\
can be sampled with no loss of information \\ using a sampling frequency $F_s \ge 2F_N$ (i.e. a sampling period $T_s \le 1/(2F_N)$)
\vspace{3em}
$F_N$ is called the Nyquist frequency of the signal.
\end{frame}
\end{document}

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